(x + 1) (x - 3) > 0, x \(\in\) R
⇒ x + 1 > 0 & x - 3 > 0
or x + 1 < 0 & x - 3 < 0
(\(\because\) if ab > 0 then either a, b > 0 or a, b < 0)
⇒ x > - 1 & x > 3 or x < - 1 & x < 3
⇒ x > 3 or x < -1
⇒ x \(\in\) (3, \(\infty\)) or x \(\in\) (-\(\infty\), -1)
⇒ x \(\in\) (-\(\infty\), -1) U (3, \(\infty\))
Alternative:
(x + 1) (x - 3) > 0
\(\therefore\) x \(\in\) (-\(\infty\), -1) U (3, \(\infty\))